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Let $z_1$ and and $z_2$ are two random variables, each having $\mathcal{N}(0,\sigma^2)$ and $\rho$ is the correlation coefficient between them. Now,

Lets $A=\min \left( z_1^2+z_2^2, \eta_0 \right)$ and $B=\min \left( z_1^2-2\rho z_1z_2+z_2^2, \eta_1 \right)$

where $\eta_0$ and $\eta_1$ are constants and $\eta_1>\eta_0$.

For asymptotic analysis, what will be the

$\mathbb{E}[AB]$ = ??

The answer of this question helped me in some way but not fully to calculate. Moments of min of a random variable and a constant.

Attempt -I:Let $\hat{A} = z_1^2+z_2^2 $ where $\mathbb{E}[\hat{A}]=\mu_A$, $var[\hat{A}]=\sigma_A^2$. Similarly $\hat{B} = z_1^2-2\rho z_1z_2+z_2^2$ where $\mathbb{E}[\hat{B}]=\mu_B$ and $var[\hat{B}]=\sigma_B^2$. Also assume $ \alpha_0 = \frac{{\eta _0 - \mu _A}}{{\sigma _A }}$, $ \alpha_1 = \frac{{\eta _1 - \mu _A}}{{\sigma _A }}$ and $\beta = \frac{{\eta _1 - \mu _A }}{{\sigma _A }} $ \begin{align} \mathbb{E}[AB]=& \mathbb{E}[AB\{\hat{A} \leq \eta_0 \}] + \eta_0 E[B\{\hat{A}> \eta_0\}] \\ =& \mathbb{E}[\hat{A}\hat{B}] \mathbb{P}\{\hat{A} \leq \eta_0 \} + \eta_0 E[B\mid \hat{A}> \eta_0] (1-\mathbb{P}\{\hat{A}\leq \eta_0\})\\ =&\mathbb{E}[\hat{A}\hat{B}]\Phi(\alpha_0)+\eta_0 E[B\mid \hat{A}> \eta_0](1-\Phi(\alpha_0)), \end{align} We can easily calculate $\mathbb{E}[\hat{A}\hat{B}]$. Now the challenging task is to calculate $E[B|\hat{A}> \eta_0]$ \begin{align} \mathbb{E}[B|\hat{A}> \eta_0]=& \mathbb{E}[B\{\eta_0<\hat{A}<\eta_1 \}] + \eta_1 E[1\{\hat{A}> \eta_1\}] \\ =& \mathbb{E}[B\mid \eta_0<\hat{A}<\eta_1 ] \mathbb{P}\{\eta_0<\hat{A}<\eta_1\} + \eta_1 \mathbb{P}\{\hat{A}> \eta_1\} \end{align} how should I calculate $\mathbb{E}[B\mid \eta_0<\hat{A}<\eta_1 ]$ =??

Attempt -II:\begin{align} \mathbb{E}[AB]=& \mathbb{E}[AB\{\hat{A} \leq \eta_0 ,\hat{B} \leq \eta_1 \}] + \mathbb{E}[AB\{\hat{A} > \eta_0 ,\hat{B} \leq \eta_1 \}]] + \mathbb{E}[AB\{\hat{A} > \eta_0 ,\hat{B} > \eta_1 \}] \\ =& \mathbb{E}[\hat{A}\hat{B}]\mathbb{P}\{\hat{A} \leq \eta_0 ,\hat{B} \leq \eta_1 \} + \eta_0\mathbb{E}[\hat{B}]\mathbb{P}\{\hat{A} > \eta_0 ,\hat{B} \leq \eta_1 \} + \eta_0\eta_1\mathbb{P}\{\hat{A} > \eta_0 ,\hat{B} > \eta_1 \} \\ \end{align} If I am correct then how should I calculate the three probabilities $\mathbb{P}\{\hat{A} \leq \eta_0 ,\hat{B} \leq \eta_1 \}$,$\mathbb{P}\{\hat{A} > \eta_0 ,\hat{B} \leq \eta_1 \}$ and $\mathbb{P}\{\hat{A} > \eta_0 ,\hat{B} > \eta_1 \}$ ??

Attempt -III: \begin{align} \mathbb{E}[AB]=& \mathbb{E}[AB\{\hat{A} \leq \eta_0 \}] + \mathbb{E}[AB\{\eta_0 < \hat{A} \leq \eta_1\}]] + \mathbb{E}[AB\{\hat{A} > \eta_1 \}] \\ \nonumber =& \mathbb{E}[\hat{A}\hat{B}]\mathbb{P}\{\hat{A} \leq \eta_0 \} + \eta_0\mathbb{E}[\hat{B}]\mathbb{P}\{\eta_0 < \hat{A} \leq \eta_1 \} + \eta_0\mathbb{E}[B \mid \hat{A}>\eta_1]\mathbb{P}\{\hat{A} > \eta_1 \} \\ \nonumber =& \mathbb{E}[\hat{A}\hat{B}]\Phi(\alpha_0) + \eta_0\mathbb{E}[\hat{B}](\Phi(\alpha_1)-\Phi(\alpha_0)) + \eta_0\mathbb{E}[B \mid \hat{A}>\eta_1](1-\Phi(\alpha_0)) \\ \nonumber \end{align} Now we need to calculate $\mathbb{E}[B \mid \hat{A}>\eta_1]$ \begin{align} \mathbb{E}[B \mid \hat{A}>\eta_1] =& \mathbb{E}[B \{\hat{B} \leq \eta_1\}] + \mathbb{E}[B \{\hat{B} > \eta_1\}]\\ \nonumber =& \mathbb{E}[\hat{B}] \mathbb{P}\{\hat{B} \leq \eta_1\} + \eta_1\mathbb{P}\{\hat{B} > \eta_1\} \\ \nonumber =& \mathbb{E}[\hat{B}] \Phi(\beta)+ \eta_1 (1-\Phi(\beta)) \end{align} From above two equations, we get \begin{align} \mathbb{E}[AB]=& \mathbb{E}[\hat{A}\hat{B}]\Phi(\alpha_0) + \eta_0\mathbb{E}[\hat{B}](\Phi(\alpha_1)-\Phi(\alpha_0)) + \eta_0(\mathbb{E}[\hat{B}] \Phi(\beta)+ \eta_1 (1-\Phi(\beta))(1-\Phi(\alpha_0)) \\ \nonumber \end{align}

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Marcus
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  • @Did Thanks. I have to mention it clearly but I had do mentioned about it. that I am finding expectation through asymptotic analysis. Here, asymptotic means if I take large number for samples of $z_1$ and $z_2$ then then $\hat{A}$ and $\hat{B}$ will follow normal distribution. – Marcus Nov 14 '16 at 10:48
  • This does not show in the definition of A and B in the question, actually it is contradictory with it. Please explain much more clearly your problem. – Did Nov 14 '16 at 10:53
  • @Did Thanks for suggestion. I have mentioned $A$ as separate variable and function of $\hat{A}$. Similarly for $B$ and $\hat{B}$. I don't say $A$ and $B$ are normally distributed but $\hat{A}$ and $\hat{B}$. It would be helpful if you can say how they are contradicting so that I can re-framed my question in better and clear way. – Marcus Nov 14 '16 at 11:07
  • Yes, and as already mentioned, neither $\hat A$ nor $\hat B$ is normal. To go further, you could explain why you think they are. Note for instance that $P(\hat A>0)=1$, a fact which suffices to show that $\hat A$ is not normal. – Did Nov 14 '16 at 11:25
  • @Did Thanks. I really appreciate your help. Actually it is a part of question that I am stuck with. Now it seems my full approach is wrong. So I decided to ask complete question so that it will be more clear. please help me to answer. http://math.stackexchange.com/questions/2013475/evaluate-mean-and-variance-of-min-function – Marcus Nov 14 '16 at 13:06